$\overline{AC}$ is $9$ units long $\overline{BC}$ is $8$ units long $\overline{AB}$ is $\sqrt{145}$ units long What is $\cos(\angle BAC)$ ? $A$ $C$ $B$ $9$ $8$ $\sqrt{145}$
SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{AC} = 9$ hypotenuse $= \overline{AB} = \sqrt{145}$ $\cos(\angle BAC )=\frac{9}{\sqrt{145}}$ $=\dfrac{9\sqrt{145} }{145}$